Answer
$\frac{\pi}{18}$
Work Step by Step
$\int_{\sqrt 3}^{3}\frac{1}{x\sqrt (4x^{2}-9)}dx$
$\int\frac{du}{u\sqrt (u^{2}-a^{2})}=\frac{1}{a}arcsec(\frac{|u|}{a})+C$
$u=2x$
$du=2dx$
$a=3$
$\frac{2}{2}\int_{\sqrt 3}^{3}\frac{1}{x\sqrt (4x^{2}-9)}dx$
$\int_{\sqrt 3}^{3}\frac{2}{2x\sqrt (4x^{2}-9)}dx$
$[\frac{1}{3}arcsec(\frac{|2x|}{3})]_{\sqrt 3}^{3}$
$\frac{1}{3}[arcsec(\frac{|2(3)|}{3})-arcsec(\frac{|2(\sqrt 3)|}{3})]$
$\frac{1}{3}[arcsec2-arcsec(\frac{2(\sqrt 3)}{3})]$
$\frac{1}{3}[\frac{\pi}{3}-\frac{\pi}{6}]$
$\frac{\pi}{18}$
